Question

If in a code language, “PAPER” is written as “QBQFS”, then “MANGO” is written as:

• A. NBOHP
• B. NBNHP
• C. NBOGP
• D. NBPHO

Hint:

When a pattern is a simple $+1$ shift, you don't need to write out every step! Just look at the first and last letters of your word: M goes to N, and O goes to P. Looking at the options, only (a) starts with N and ends with P!

Answer 1

The Correct Option is A

Step 1 : Understanding the Question:
The topic of this question is Logical Reasoning, specifically focusing on the sub-topic of Coding-Decoding. In these types of problems, a specific rule or transformation is applied to a word to convert it into a code. The student must identify the pattern by analyzing the relationship between the letters of the given word and its corresponding code. Once the logic is established, the same rule must be applied to a new target word to find its coded version among the given options.
Step 2 : Key Formulas and approach:
The approach for solving letter-based coding involves identifying the alphabetical displacement (shift) of each character. If we represent the position of a letter in the English alphabet as $P$ (where $A=1, B=2, \dots$), the transformation rule can be expressed as:
$$\text{Coded Letter} = \text{Position}(P + n)$$
Where $n$ represents the numerical shift. For this problem, the approach is:
1. Map each letter of the word "PAPER" to its corresponding letter in "QBQFS".
2. Determine if the shift $n$ is constant or variable.
3. Apply the identified shift to the letters of the word "MANGO" sequentially.
Step 3 : Detailed Explanation:

We start by examining the word "PAPER" and its code "QBQFS" to find the underlying logic. Let's compare them letter by letter:

The first letter 'P' in the original word is transformed into 'Q'. In the English alphabet, 'Q' comes immediately after 'P', indicating a shift of $+1$.

The second letter 'A' is transformed into 'B'. Since 'B' is the next letter after 'A', the shift remains $+1$.

The third letter 'P' is again transformed into 'Q', maintaining the $+1$ alphabetical displacement.

The fourth letter 'E' is transformed into 'F'. Counting one step forward from 'E' gives 'F', confirming the $+1$ shift.

The fifth letter 'R' is transformed into 'S'. Following the pattern, 'S' is the letter immediately following 'R', completing the $+1$ logic for the entire word.

Now that we have established the rule is a constant $+1$ shift for every character, we apply it to the target word "MANGO".

For 'M': The next letter in the alphabet is 'N'.

For 'A': The next letter in the alphabet is 'B'.

For 'N': The next letter in the alphabet is 'O'.

For 'G': The next letter in the alphabet is 'H'.

For 'O': The next letter in the alphabet is 'P'.

Combining these results, the coded word for "MANGO" becomes "NBOHP".

Looking at the options provided: (A) NBOHP, (B) NBNHP, (C) NBOGP, and (D) NBPHO, only option (A) perfectly matches our derived code.

Step 4 : Final Answer:
By applying the constant +1 alphabetical shift to each letter of the word "MANGO," we obtain the code NBOHP, which is Option (A).